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How to determine the resistance in a electric heater

  1. Sep 15, 2011 #1
    hallo everyone,

    I have a problem which i cant solve. Does somebody know how to solve this problem?? The question is:

    A 220 V electric heater has two heating coils which can be switched such that either coil can be used independently or can be connected in series or parallel, yielding a total of four possible configurations.
    The warmest setting corresponds to 2000 W and the coolest corresponds to 300 W.
    R1 > R2

    Determine the resistance of R1 [in Ω].

    regards
     
  2. jcsd
  3. Sep 15, 2011 #2

    Hootenanny

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    What have you tried so far? How do you think you can approach it?
     
  4. Sep 15, 2011 #3
    I've drawn four configurations;
    1) two resistors parallel (R1 and R2) with a source voltage of 220 V
    2) 1 resistor with a source voltage of 220 V
    3) 1 resistor with a source voltage of 220 V
    4) two resistors (R1 and R2) in series with a source voltage of 220 V

    For each of these configurations I've computed the equivalent resistance and concluded that configuration 1 consumes the most power (so 2000 W) and config. 2 the least power (300 W). So for config. 1 I have 4 unknowns; R1, R2, I1, I2 and I have 3 equations; R1=U/I1 R2=U/I2 and finally P=(I1^2)*R1 + (I2^2)*R2. So I'm missing one equation to solve this problem and I do not know which one. I've also tried to use the last configuration and to relate these configs but I didn't get an answer that makes reason.
     
  5. Sep 15, 2011 #4

    Hootenanny

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    You just need to think a little more. For example, in the parallel circuit you know that [itex]V_1=V_2=V[/itex]. Thus,

    [tex]P_p = V^2\left(\frac{1}{R_1} + \frac{1}{R_2}\right)[/tex]

    And you already know the parallel circuit power consumption. Now, how about the single resistor circuit?
     
  6. Sep 15, 2011 #5

    gneill

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    Wouldn't the least power occur when the net resistance is as large as possible? After all, P = V2/R.
     
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